Question

Aflati numerele naturale care impartindule la 9 dau catul si restul doua numere naturale consecutive.

Answer 1

n = 9c + r        r < 9        r ∈ {0,1,2,.....8}
|.  c = r - 1  ⇒ c ∈ {0,1,2,......7}⇒  n ∈ {1,11,21,31,....71}
||. c = r + 1 ⇒  c ∈ {1,2,3.......9} ⇒ n ∈ {9,19,29,39......89}

Answer 2

n:9=0 rest 1
n=0x9+1 ⇒ n=1
n:9=1 rest 2
n=1x9+2 ⇒ n=11
n:9=2 rest 3
n=2x9+3 ⇒ n=21
n:9=3 rest 4
n=3x9+4 ⇒ n=31
n:9=4 rest 5
n=4x9+5 ⇒ n=41
n:9=5 rest 6
n=5x9+6 ⇒ n=51
n:9=6 rest 7
n=6x9+7 ⇒ n=61
n:9=7 rest 8
n=7x9+8 ⇒ n=71
R: nr. sunt :  1 ; 11 ; 21 ; 31 ; 41 ; 51 ; 61 ; 71